./run.sh:
1. cp=warmonger-1.0.0.jar
2. cmmd="java -server -D64 -Xms200m -Xmx200m
-Dlog4j.configurationFile=$WARMONGER_HOME/etc/log4j2.xml
-classpath $cp warmonger.agent.WarmongerAgentApp"
3. echo $cmmd
execute results:
dataq.agent.DataqAgentApp -Xmx200m
-Dlog4j.configurationFile=/warmonger/etc/log4j2.xml
-classpath warmonger-1.0.0.jar
"warmonger.agent.WarmongerAgentApp" not appear.
I means if remove echo, java will be throw an exception: Couldn't find main class
You won't see $cp when you echo $cmmd, because the shell substitutes the value of the cp parameter (warmonger-1.0.0.jar) in the assignment to cmmd.
You can escape the dollar sign, or use single quotes if you don't want the shell to expand the parameter.
Your shell script is CR-LF terminated (DOS/Windows ends of lines). Thus, from bash point of view, the cp variable contains warmonger-1.0.0.jar<CR> (notice the trailing <CR>).
When you echo the content of the cp variable, <CR> is echoed too which puts the cursor at the beginning of the line (CR = carriage return). echo then prints the remaining of the arguments at the beginning of the line.
You can see it in your output:
"java -server -D64 -Xms200m -Xmx200m" is overwritten with "warmonger.agent.WarmongerAgentApp"
which, in turn, is overwritten by some other command output ("dataq.agent.Dataq")
Solution: turn your DOS/Windows text file into a UNIX one. See this answer.
Related
My python script can take a series of bitwise operators as one of its arguments. They all work fine except for "=<<" which is roll left, and "=>>" which is roll right. I run my script like ./script.py -b +4,-4,=>>10,=<<1, where anything after -b can be any combination of similar operations. As soon as the terminal sees "<<" though, it just drops the cursor to a new line after the command and asks for more input instead of running the script. When it sees ">>", my script doesn't process the arguments correctly. I know it's because bash uses these characters for a specific purpose, but I'd like to get around it while still using "=>>" and "=<<" in my arguments for my script. Is there any way to do it without enclosing the argument in quotation marks?
Thank you for your help.
You should enclose the parameters that contain special symbols into single quotation marks (here, echo represents your script):
> echo '+4,-4,=>>10,=<<1'
+4,-4,=>>10,=<<1
Alternatively, save the parameters to a file (say, params.txt) and read them from the file onto the command line using the backticks:
> echo `cat params.txt`
+4,-4,=>>10,=<<1
Lastly, you can escape some offending symbols:
> echo +4,-4,=\>\>10,=\<\<1
+4,-4,=>>10,=<<1
I have a Bash script which generates, stores and modifies values in an array. These values are later used as arguments for a command.
For a MCVE I thought of an arbitrary command bash -c 'echo 0="$0" ; echo 1="$1"' which explains my problem. I will call my command with two arguments -option1=withoutspace and -option2="with space". So it would look like this
> bash -c 'echo 0="$0" ; echo 1="$1"' -option1=withoutspace -option2="with space"
if the call to the command would be typed directly into the shell. It prints
0=-option1=withoutspace
1=-option2=with space
In my Bash script, the arguments are part of an array. However
#!/bin/bash
ARGUMENTS=()
ARGUMENTS+=('-option1=withoutspace')
ARGUMENTS+=('-option2="with space"')
bash -c 'echo 0="$0" ; echo 1="$1"' "${ARGUMENTS[#]}"
prints
0=-option1=withoutspace
1=-option2="with space"
which still shows the double quotes (because they are interpreted literally?). What works is
#!/bin/bash
ARGUMENTS=()
ARGUMENTS+=('-option1=withoutspace')
ARGUMENTS+=('-option2=with space')
bash -c 'echo 0="$0" ; echo 1="$1"' "${ARGUMENTS[#]}"
which prints again
0=-option1=withoutspace
1=-option2=with space
What do I have to change to make ARGUMENTS+=('-option2="with space"') work as well as ARGUMENTS+=('-option2=with space')?
(Maybe it's even entirely wrong to store arguments for a command in an array? I'm open for suggestions.)
Get rid of the single quotes. Write the options exactly as you would on the command line.
ARGUMENTS+=(-option1=withoutspace)
ARGUMENTS+=(-option2="with space")
Note that this is exactly equivalent to your second option:
ARGUMENTS+=('-option1=withoutspace')
ARGUMENTS+=('-option2=with space')
-option2="with space" and '-option2=with space' both evaluate to the same string. They're two ways of writing the same thing.
(Maybe it's even entirely wrong to store arguments for a command in an array? I'm open for suggestions.)
It's the exact right thing to do. Arrays are perfect for this. Using a flat string would be a mistake.
I am trying to create a shell script for logs and trying to append data into a text file. I have write this sample "test.sh" code for testing:
#!/bin/sh -e
touch /home/sample.txt
SPTH = '/home/sample'.txt
echo "MY LOG FILE" >> "$SPTH"
echo "DUMP started at $(date +'%d-%m-%Y %H:%M:%S')" >> /home/sample.txt
echo "DUMP finished at $(date +'%d-%m-%Y %H:%M:%S')" >> /home/sample.txt
but in above code all lines are working correct except one line of code i.e.
echo "MY LOG FILE" >> "$SPTH"
It is giving error:
test.sh: line 6: : No such file or directory
I want to replace this full path of file "/home/sample.txt" to variable "$SPATH".
I am executing my shell script using
sh test.sh
What I am doing wrong.
Variable assignments in bash shell does not allow you to have spaces within. It will be actually interpreted as command with = and the subsequent keywords as arguments to the first word, which is wrong.
Change your code to
SPTH="/home/sample.txt"
That is the reason why SPTH was not assigned to the actual path you intended it to have. And you have no reason to have single-quote here and excluding the extension part. Using it fully within double-quotes is absolutely fine.
The syntax for the command line is that the first token is a command, tokens are separated by whitespace. So:
SPTH = '/home/sample'.txt
Has the command as SPTH, the second token is =, and so on. You might think this is daft, but most shells behave like this for historical reasons.
So you need to remove the whitespace:
SPTH='/home/sample'.txt
I have a problem with "/" sign in bash shell (version 4.3 in Ubuntu 16). I have a function:
command_not_found_handler() {
if [[ "$1" =~ any$ ]]; then
echo "$1"
fi
}
This function should write back the contents of any command written in the terminal when this command ends with any.
This works well, except in the situation when I write something with a /, such as whatever/any. In that event, I receive an error akin to the following:
bash: no such file or directory: whatever/any
Any attempts to escape this / in the function have no effect (for instance if [[ "$1" =~ /any$ ]]; then or if [[ "$1" =~ \/any$ ]]; then).
What can I do to make it work with / sign?
command_not_found_handle (no trailing r) is only invoked after doing a search through the PATH for a given command.
No such search occurs when the user is passing an explicit path to a command, which is how anything containing a / is interpreted.
To quote the relevant documentation, with emphasis added:
If the name is neither a shell function nor a builtin, and contains no slashes, Bash searches each element of $PATH for a directory containing an executable file by that name. Bash uses a hash table to remember the full pathnames of executable files to avoid multiple PATH searches (see the description of hash in Bourne Shell Builtins). A full search of the directories in $PATH is performed only if the command is not found in the hash table. If the search is unsuccessful, the shell searches for a defined shell function named command_not_found_handle. If that function exists, it is invoked with the original command and the original command’s arguments as its arguments, and the function’s exit status becomes the exit status of the shell. If that function is not defined, the shell prints an error message and returns an exit status of 127.
The entire paragraph of documentation is relevant only in the set of conditions set out at the beginning: A command must not be a shell function, must not be a builtin, and must not contain slashes.
I have a test script which has a lot of commands and will generate lots of output, I use set -x or set -v and set -e, so the script would stop when error occurs. However, it's still rather difficult for me to locate which line did the execution stop in order to locate the problem.
Is there a method which can output the line number of the script before each line is executed?
Or output the line number before the command exhibition generated by set -x?
Or any method which can deal with my script line location problem would be a great help.
Thanks.
You mention that you're already using -x. The variable PS4 denotes the value is the prompt printed before the command line is echoed when the -x option is set and defaults to : followed by space.
You can change PS4 to emit the LINENO (The line number in the script or shell function currently executing).
For example, if your script reads:
$ cat script
foo=10
echo ${foo}
echo $((2 + 2))
Executing it thus would print line numbers:
$ PS4='Line ${LINENO}: ' bash -x script
Line 1: foo=10
Line 2: echo 10
10
Line 3: echo 4
4
http://wiki.bash-hackers.org/scripting/debuggingtips gives the ultimate PS4 that would output everything you will possibly need for tracing:
export PS4='+(${BASH_SOURCE}:${LINENO}): ${FUNCNAME[0]:+${FUNCNAME[0]}(): }'
In Bash, $LINENO contains the line number where the script currently executing.
If you need to know the line number where the function was called, try $BASH_LINENO. Note that this variable is an array.
For example:
#!/bin/bash
function log() {
echo "LINENO: ${LINENO}"
echo "BASH_LINENO: ${BASH_LINENO[*]}"
}
function foo() {
log "$#"
}
foo "$#"
See here for details of Bash variables.
PS4 with value $LINENO is what you need,
E.g. Following script (myScript.sh):
#!/bin/bash -xv
PS4='${LINENO}: '
echo "Hello"
echo "World"
Output would be:
./myScript.sh
+echo Hello
3 : Hello
+echo World
4 : World
Workaround for shells without LINENO
In a fairly sophisticated script I wouldn't like to see all line numbers; rather I would like to be in control of the output.
Define a function
echo_line_no () {
grep -n "$1" $0 | sed "s/echo_line_no//"
# grep the line(s) containing input $1 with line numbers
# replace the function name with nothing
} # echo_line_no
Use it with quotes like
echo_line_no "this is a simple comment with a line number"
Output is
16 "this is a simple comment with a line number"
if the number of this line in the source file is 16.
This basically answers the question How to show line number when executing bash script for users of ash or other shells without LINENO.
Anything more to add?
Sure. Why do you need this? How do you work with this? What can you do with this? Is this simple approach really sufficient or useful? Why do you want to tinker with this at all?
Want to know more? Read reflections on debugging
Simple (but powerful) solution: Place echo around the code you think that causes the problem and move the echo line by line until the messages does not appear anymore on screen - because the script has stop because of an error before.
Even more powerful solution: Install bashdb the bash debugger and debug the script line by line
If you're using $LINENO within a function, it will cache the first occurrence. Instead use ${BASH_LINENO[0]}